Question: There are $9$ students in a class: $7$ boys and $2$ girls. If the teacher picks a group of $3$ at random, what is the probability that everyone in the group is a boy?
One way to solve this problem is to figure out how many different groups there are of only boys, then divide this by the total number of groups you could choose. Since every group is chosen with equal probability, this will be the probability that a group of all boys is chosen. We know two ways to count the number of groups we can choose: we use permutations if order matters, and combinations if it doesn't. Does the order the students are picked matter in this case? It doesn't matter if we pick John then Ben or Ben then John, so order must not matter. So, the number of ways to pick a group of $3$ students out of $9$ is $ \dfrac{9!}{(9-3)!3!} = \binom{9}{3}$. We can use the same logic to count the number of groups that only have boys. Specifically, the number of ways to pick a group of $3$ students out of $7$ is $ \dfrac{7!}{(7-3)!3!} = \binom{7}{3}$. So, the probability that the teacher picks a group of all boys is the number of groups with only boys divided by the number of total groups the teacher could pick. This is $ \frac{\frac{7!}{(7-3)!\cancel{3!}}} {\frac{9!}{(9-3)!\cancel{3!}}} = \frac{\frac{7!}{4!}}{\frac{9!}{6!}}$ We can re-arrange the terms to make simplification easier $\left(\dfrac{7!}{4!}\right) \left(\dfrac{6!}{9!}\right) = \left(\dfrac{7!}{9!}\right) \left(\dfrac{6!}{4!}\right)$ Simplifying, we get $ \left(\dfrac{\cancel{7!}}{9\cdot8 \cdot \cancel{7!}}\right) \left(\dfrac{6\cdot5 \cdot \cancel{4!}}{\cancel{4!}}\right) = \left(\dfrac{1}{72}\right) \left(30\right) = \dfrac{5}{12}$